How many times must a man toss a fair coin, so that the probability of

1.	How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?
Answer» Let no. of times of tossing a coin be n. Here, Probability of getting a head in a chance = p = 1/2 Probability of getting no head in a chance = q = 1 - 1/2 = 1/2 Now, P (having at least one head) = P (X ≥ 1) = 1 - P(X = 0) = 1 - ⁿC₀P⁰p⁰q^{n - 0}= 1 - 1.1.(1/2)ⁿ = 1 - (1/2)ⁿ From question 1 - (1/2)ⁿ > 80/100 ⇒ 1 - (1/2)ⁿ > 8/10 ⇒ 1 - 8/10 > 1/2ⁿ ⇒ 1/5 > 1/2ⁿ ⇒ 2ⁿ > 5 ⇒ n ≥ 3 A man must have to toss a fair coin 3 times.

How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?

Answer»

Let no. of times of tossing a coin be n.

Here, Probability of getting a head in a chance = p = 1/2

Probability of getting no head in a chance = q = 1 - 1/2 = 1/2

Now, P (having at least one head) = P (X ≥ 1)

= 1 - P(X = 0) = 1 - ⁿC₀P⁰p⁰q^{n - 0}= 1 - 1.1.(1/2)ⁿ = 1 - (1/2)ⁿ

From question

1 - (1/2)ⁿ > 80/100

⇒ 1 - (1/2)ⁿ > 8/10 ⇒ 1 - 8/10 > 1/2ⁿ

⇒ 1/5 > 1/2ⁿ ⇒ 2ⁿ > 5 ⇒ n ≥ 3

A man must have to toss a fair coin 3 times.

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How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?

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